I was finally able to figure it out. I'll post it for the sake of completeness.Originally Posted by topsquark
Our story so far...
The subring R of the field K is actually an integral domain, so R contains no zero-divisors. Additionally, the hypothesis assures us that we have that and such that:
This leads to the result that has a multiplicative inverse in R.
The problem at this point is that we need to prove that an inverse exists for all elements of R. I was finally able to get around that problem:
Since , take any and form the product polynomial:
This produces a new polynomial:
By the construction in the previous post (and as repeated above) the existence of this polynomial implies that the element has an inverse in R also. Recall that multiplication in R is commutative (since it is a subring of the field K.) Thus:
, since multiplication in R is closed by definition.
So all elements of R have a multiplicative inverse, making the integral domain R a field.